By Math Vault | Calculus

Like the featured image above? Good for you! As students (fanatics?) of mathematics and other technical sciences, we have an intuitive appreciation of the role this topic — **calculus** — plays out in our daily life. While obviously not a subject to be mastered overnight, what’s true is that our first exposure to it tends to be almost universally restricted to the topic of **derivatives** of *elementary functions. *

Interestingly enough, when it comes to the *calculation of derivatives*, there is a rule of thumb out there that goes something like this: either the function is *basic*, in which case we can appeal to the **table of derivatives**, or the function is *composite*, in which case we can differentiated it *recursively* — by breaking it down into the derivatives of its *constituents* via a series of **derivative rules**.

Indeed, for those of us who have ever read a calculus textbook or attended a calculus course, we might be already familiar with most — if not all — of these derivative rules:

If the function $cf$ is defined on an interval $I$ and $f$ is differentiable on $I$, then $\displaystyle (cf)’=cf’$ on $I$.

Given a *real* number $r$ greater or equal to $1$ , $(x^r)’ = r x^{r-1}$ for all $x \in \mathbb{R}$. In the case where $r$ is less than $1$ (and *non-zero*), $(x^r)’ = r x^{r-1}$ for all $x \ne 0$.

If the function $f+g$ is well-defined on an interval $I$, with $f$ and $g$ being *both* differentiable on $I$, then $\displaystyle (f+g)’ = f’ + g’$ on $I$.

If the function $f-g$ is well-defined on an interval $I$, with $f$ and $g$ being *both* differentiable on $I$, then $\displaystyle (f-g)’ = f’ – g’$ on $I$.

If the function $fg$ is well-defined on an interval $I$, with $f$ and $g$ being *both* differentiable on $I$, then $\displaystyle (fg)’ = f’g + fg’$ on $I$.

If the function $\displaystyle \frac{1}{f}$ is well-defined on an interval $I$, with $f$ being differentiable on $I$, then $\displaystyle \left( \frac{1}{f} \right)’ = \frac{-f’}{f^2}$ on $I$.

If the function $\displaystyle \frac{f}{g}$ is well-defined on an interval $I$, with $f$ and $g$ being *both* differentiable on $I$, then $\displaystyle \left( \frac{f}{g} \right)’ = \frac{f’g – fg’}{g^2}$ on $I$.

If the function $f \circ g$ in well-defined on an interval $I$, with $g$ being differentiable on $I$ and $f$ being differentiable on $g(I)$, then:

\begin{align*} [(f \circ g)(x)]’ = f'[g(x)] \, g'(x) \qquad (x \in I) \end{align*}

(refer to the module on **Chain Rule** for more detail)

If the function $f^{-1}$ is well-defined on an interval $I$, and the expression $\displaystyle \frac{1}{ f'[ f^{-1} (x) ] }$ makes sense on $I$ (i.e., for all $x \in I$, $f$ is differentiable at $f^{-1} (x)$, with the derivative being *non-zero*), then:

\begin{align*} [f^{-1}(x)]’ = \frac{1}{f'[f^{-1}(x)]} \qquad (x \in I) \end{align*}

(refer to the module on **Inverse Function Theorem** for more detail)

For the most part, these rules are more than enough to handle the vast majority of functions one would encounter. However, when we look at our *repertoire of functions*, we see that something else is missing, namely:

## What about the functions constructed via exponentiation?

Here, prompted by an *unusual* sense of urgency, we proceed to toy around with the idea of *derivative of a power*, subsequently ending up developing a rule precisely for that purpose. The name? **Exponent Rule** of course! Now, we bet that you haven’t seen this one yet — at least not in a standard calculus textbook or Khan Academy anyway!

Given a **base function** $f$ and an **exponent function** $g$ — both defined on an interval $I$ — one can construct the **power function** $f^g$ (i.e., $f$ raised to $g$), which is defined as long as $f(x)>0$ on $I$ (why?). In which case, the **properties of logarithm** warrant that:

\begin{align*} f(x)^{g(x)} & = e^{\ln \left[ f(x)^{g(x)} \right] } = e^{g(x) \ln f(x)} \end{align*}

which suggests that $\displaystyle e^{g(x) \ln f(x)}$ and $\displaystyle f(x)^{g(x)}$ can be used interchangeably on $I$.

More importantly though, in the case where $f$ and $g$ are *both differentiable* on $I$, using a combination of **Chain Rule** and **Product Rule**, we get that:

\begin{align*} \left( f(x)^{g(x)} \right)’ & = \left( e^{g(x) \ln f(x)} \right)’ \\ & = e^{g(x) \ln f(x)} \, [g(x) \ln f(x)]’ \\ & = f(x)^{g(x)} \left[ g'(x) \ln f(x) + f'(x) \frac{ g(x)}{f(x)} \right] \end{align*}

Too much $f$ and $g$? No worry. All we have shown is that the derivative of $f^g$ can be expressed in terms of the derivatives of $f$ and $g$ — and hence the **Exponent Rule** for derivatives!

Theorem — The Exponent Rule for Derivative

Given a **base** function $f$ and an **exponent** function $g$, if:

- The
**power function**$f^g$ is*well-defined*on an interval $I$ (i.e.,$f$ and $g$ both well-defined on $I$, with $f>0$ on $I$) - Both $f$ and $g$ are
*differentiable*on $I$

then the function $f^g$ is *differentiable* on $I$ as well. In addition:

\begin{align*} (f^g)’ = f^g \left( g’ \ln f + f’ \frac{g}{f} \right) \qquad (x \in I) \end{align*}

In case it’s still not clicking, this is what the **Exponent Rule** reads in English:

## The derivative of a power, is equal to the power itself times the following: the derivative of the exponent times the logarithm of the base, plus the derivative of the base times the exponent-base ratio..

Wonderful *meditation mantra*? Or maybe a *bad-pun-turned-sour-joke*? Either way, we just want you to know that there is *definitely* a pattern to this! Indeed, the Exponent Rule basically states that the derivative of $f^g$ takes the form of $f^g (A+B)$, where:

- $A$ is obtained by taking the derivative of the
*exponent*, times the logarithm of the*base*. - $B$ is obtained by taking the derivative of the
*base*instead, times the ratio with the*exponent*on the top.

In fact, with just a bit of practice, it’s possible to master the **Exponent Rule** as much as we do with the *Quotient Rule* — and this is not to mention the whole new world it opens up to our *mental calculus fanatics*!

Traditionally, to evaluate the derivative of a *power function*, one would have to either resort to **logarithmic differentiation**, or **base-e standardisation** before differentiating it. With the advent of the Exponent Rule, both of these approaches basically become obsolete — not because they are irrelevant per se, but because they have already been carried out *during* the derivation of the Exponent Rule.

In what follows, we lay out 7 examples illustrating how the Exponent Rule can be used to *streamline* the differentiation process for **power functions**, thereby freeing up some of our brainpower and resources to other potentially-more-challenging tasks that are yet to come.

With the base function being $\pi$ and the exponent function $x$, the exponential function $\pi^x$ — which is defined on the *entire* real line — easily satisfies the *preconditions* laid out by the Exponent Rule. In which case, differentiating it using the Rule yields that:

\begin{align*} (\pi^x)’ & = \pi^x \left[ (x)’ \ln \pi + (\pi)’ \frac{x}{\pi} \right] = \pi^x \ln \pi \qquad (x \in \mathbb{R})\end{align*}

which shows that $\pi^x$ differentiates to itself along with some *adjustment factor* — a typical behavior among the exponential functions.

Now, is that a *heavy-handed* differentiation approach to a particularly-simple function? Actually, not really, for as it turns out, the alternative approaches would have been essentially the same — save perhaps a few more extra *“patching” steps*.

Finally, to celebrate our first *successful* execution of the Exponent Rule, we’ve cooked up the graphs of $\pi^x$ and its derivative here:

Generalizing from the last example, we see that given the base function $b$ ($b>0, b \ne 1$) and the exponent function $x$, the **exponential function** $b^x$ — which is defined on the *entire* real line — can also be differentiated using the Exponent Rule as follows:

\begin{align*} (b^x)’ & = b^x \left[ (x)’ \ln b + (b)’ \frac{x}{b} \right] = b^x \ln b \qquad (x \in \mathbb{R})\end{align*}

which shows that in general, an exponential function differentiates to itself, along with the *natural logarithm* of its base.

With the base function $x$ and the exponent function $\ln x$, the power function $\displaystyle x^{\ln x}$ unfortunately can be only defined on the set of *positive* real numbers $\mathbb{R_+}$. However, this is really more of a blessing than a curse, as $\mathbb{R_+}$ happens to be the *biggest* domain under which both the base and the exponent function are differentiable anyway. In which case, applying the **Exponent Rule** yields that:

\begin{align*} (x^{\ln x})’ & = x^{\ln x} \left[ (\ln x)’ (\ln x) + (x)’\frac{\ln x}{x} \right] \\ & = x^{\ln x} \left[ \frac{\ln x}{x} + \frac{\ln x}{x} \right] \\ & = x^{\ln x} \, \frac{2 \ln x}{x} \qquad (x>0) \end{align*}

which is not too bad as a result. To be sure, here’s the graph of $\displaystyle x^{\ln x}$ and its derivative alongside with it:

There we go. Time to unleash the e*xtravaganza*! With the base function $x^2+1$ and the exponent function $\sin x$, the power function $\displaystyle (x^2+1)^{\sin x}$ — which is defined *everywhere* on $\mathbb{R}$ — has the added bonus that both the base and the exponent functions are differentiable on $\mathbb{R}$ as well. In which case, a single application of the **Exponent Rule** yields:

\begin{align*} \left( (x^2+1)^{\sin x}\right)’ & = (x^2+1)^{\sin x} \left[ (\sin x)’ \ln (x^2+1) + (x^2+1)’ \frac{\sin x}{x^2+1} \right] \\ & = (x^2+1)^{\sin x} \left[ \cos x \ln (x^2+1) + \frac{2x \sin x}{x^2+1} \right] \end{align*}

Interesting! Let’s plug them into our graphing application and see what kind of *madness* it produces in return!

Running through our checklist as usual, we see that the base function $2x$ can only be *positive* on $\mathbb{R_+}$ (i.e., the set of *positive* real numbers). However, in doing so, we also found that both the base function $2x$ and the exponent function $3x$ are *well-defined* and *differentiable* on $\mathbb{R_+}$. As a result, the power function $\displaystyle (2x)^{3x}$ is differentiable on $\mathbb{R_+}$ as well, with its derivative obtainable through the use of **Exponent Rule**:

\begin{align*} \left[ (2x)^{3x}\right]’ & = (2x)^{3x} \left[ (3x)’ \ln (2x) + (2x)’ \frac{3x}{2x} \right] \\ & = (2x)^{3x} \left[ 3 \ln(2x) + 3 \right] \qquad (x>0) \end{align*}

Alternatively, we can also push the $3$ in the exponent function into the base function, as follows:

\begin{align*} (2x)^{3x} & = \left[{(2x)^3}\right]^x = (8x^3)^x \qquad (x>0)\end{align*}

in which case, the previous observations about the *domain* and *differentiability* still hold, and applying the Exponent Rule once again yields:

\begin{align*} \left[ (8x^3)^x \right]’ & = (8x^3)^x \left[ \ln (8x^3) + 24x^2 \frac{x}{8x^3} \right] \\ & = (2x)^{3x} \left[ \ln (8x^3) + 3 \right] \qquad (x>0) \end{align*}

which coincides with the derivative obtained a bit earlier. Magic!

So, what about the cases involving *higher exponentiation* like this one? Well, no sweat here, as this is not a real *nested* power function anyway. Indeed, using the **Generalized Power Rule for Logarithm**, one can *collapse* it back into an *ordinary* power function as follows:

\begin{align*} {\left( \sqrt{x}^{\cos x}\right)}^{\ln x} & = \sqrt{x}^{\cos x \ln x} \qquad (x>0) \end{align*}

Once here, it can be seen that not only is $\displaystyle \sqrt{x}^{\, \cos x \ln x}$ well-defined on $\mathbb{R_+}$, but both the base function $\sqrt{x}$ and the exponent function $\cos x \ln x$ are differentiable on $\mathbb{R_+}$ as well. In which case, applying the **Exponent Rule** yields that:

\begin{align*} \left( \sqrt{x}^{\, \cos x \ln x} \right)’ & = \sqrt{x}^{\, \cos x \ln x} \left[ \left( -\sin x \ln x + \frac{\cos x}{x} \right) \ln \sqrt{x} + \frac{1}{2\sqrt{x}} \frac{\, \cos x \ln x}{\sqrt{x}} \right] \\ & = \sqrt{x}^{\, \cos x \ln x} \left[ -\frac{1}{2}\sin x (\ln x)^2 + \frac{\cos x \ln x}{x} \right] \qquad (x>0) \end{align*}

which gets a bit lengthy. But the point is that it is doable, so that if we put everything together, we get that:

\begin{align*} \left[ {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \right]’ & = \left( \sqrt{x}^{\, \cos x \ln x} \right)’ \\ & = {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \left[ -\frac{1}{2}\sin x (\ln x)^2 + \frac{\cos x \ln x}{x} \right] \qquad (x>0) \end{align*}

and here’s a picture of the graph as a bonus.

Now you might be asking, “isn’t that the one we’ve just worked on a bit earlier?” Actually, not quite. This is due to the *not-so-secret* convention that:

\begin{align*} \sqrt{x}^{\, \cos x^{\, \ln x}} & \stackrel{df}{=} \sqrt{x}^{\, (\cos x^{\, \ln x} ) } \end{align*}

In this case, the **Power Rule for logarithm** simply won’t work. But then, that’s not to say that it precludes us from calculating the derivative *recursively* either.

To see how, let’s begin by focusing on the exponent function $\displaystyle \cos x^{\, \ln x}$, from which we can see that:

- $\ln x$ is only defined when $x>0$.
- $\cos x>0$ only when $\displaystyle -\frac{\pi}{2} < x < \frac{\pi}{2} \pmod{2\pi}$.

So that if we let $I$ denote the set satisfying *both* restrictions. That is,

\begin{align*}I \stackrel{df}{=} \left\{ x \in \mathbb{R_+} \mathrel{\Big|} -\frac{\pi}{2} < x < \frac{\pi}{2} \,\, (\text{mod } 2\pi ) \right\} \end{align*}

then one can see that $\displaystyle \cos x^{\, \ln x}$ is well-defined on $I$, with both $\cos x$ and $\ln x$ being differentiable on $I$ as well. In which csae, the **Exponent Rule** kicks in, yielding that:

\begin{align*} \left( \cos x^{\, \ln x} \right)’ & = \cos x^{\, \ln x} \left[ \frac{1}{x} \ln (\cos x) + (-\sin x) \frac{\ln x}{\cos x} \right] \\ & = \cos x^{\, \ln x} \left[ \frac{\ln (\cos x)}{x} – \tan x \ln x \right] \qquad (x \in I) \end{align*}

which takes care of the derivative of the *exponent function*. Now, if we just backtrack a bit to the *original function*, then it shouldn’t be hard to to see that:

- $\displaystyle \sqrt{x}^{\, (\cos x^{\, \ln x})}$ is well-defined on $I$.
- Both the base function $\sqrt{x}$ and the exponent function $\cos x^{\, \ln x}$ are differentiable on $I$.

which can only mean one thing — that $\displaystyle \sqrt{x}^{\, (\cos x^{\ln x})}$ is very well differentiable on $I$ as well. Here, applying the **Exponent Rule **a second time, we obtain that:

\begin{align*} \left[ \sqrt{x}^{\, ({\cos x}^{\, \ln x} )} \right]’ & = \sqrt{x}^{\, ({\cos x}^{\, \ln x} )} \left[ \left( {\cos x}^{\, \ln x}\right)’ \ln \sqrt{x} + \frac{1}{2\sqrt{x}} \frac{{\cos x}^{\, \ln x}}{\sqrt{x}} \right] \\ & = \cdots \end{align*}

Well, guess we’ll stop here and leave the algebra up to you. But hey, if you are proficient enough to have gone this far, you *can* finish that up right? 🙂

And for your curiosity, here’s the graph of the original function and its derivative for the record. Notice the *stark contrast* between this and the graphs from Example 6.

Ouf! That was a bit of symbol-crunching up there! But hopefully it illustrates why the Exponent Rule can be a valuable asset in our arsenal of **derivative rules**. While for simple power function, this approach might seem like an *overkill*, for *repeatedly-exponentiated* power functions with one *nested* inside another, it becomes readily apparent that the **Exponent Rule** is absolutely the way to go.

In addition to *automating* the differentiation process for power functions, the Exponent Rule — especially when *combined* with other **traditional derivative rules** — can really do wonder in terms of tapping into functions that are previously too *intimidating* / *tedious* to tackle — such as the ones we have a hard time finding in a typical calculus textbook for instance. 🙂

In any case, let’s leave it at that and call it a day for now. By the way, here’s an *interactive table* summarizing what we have discovered thus far — for the sake of completeness:

Constant Rule |

Power Rule |

Sum Rule |

Difference Rule |

Product Rule |

Reciprocal Rule |

Quotient Rule |

Chain Rule |

Inverse-Function Rule |

Given a **base** function $f$ and an **exponent** function $g$, if:

- The
**power function**$f^g$ is*well-defined*on an interval $I$ (i.e.,$f$ and $g$ both well-defined on $I$, with $f>0$ on $I$) - Both $f$ and $g$ are
*differentiable*on $I$.

then the function $f^g$ is differentiable on $I$ as well. In addition:

\begin{align*} (f^g)’ = f^g \left( g’ \ln f + f’ \frac{g}{f} \right) \qquad (x \in I) \end{align*}

$\displaystyle (\pi^x)’ = \pi^x \ln \pi \quad (x \in \mathbb{R})$ |

$\displaystyle (b^x)’ = b^x \ln b \quad (x \in \mathbb{R})$ |

$\displaystyle (x^{\ln x})’ = x^{\ln x} \, \frac{2 \ln x}{x} \quad (x>0) $ |

$\displaystyle \left[ (x^2+1)^{\sin x} \right]’ = (x^2+1)^{\sin x} \left[ \cos x \ln (x^2+1) + \frac{2x \sin x}{x^2+1} \right] \quad (x \in \mathbb{R}) $ |

$\displaystyle \left[ (2x)^{3x}\right]’ = (2x)^{3x} \left[ 3 \ln(2x) + 3 \right] \quad (x>0) $ |

$\displaystyle \left[ {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \right]’ = {\left( \sqrt{x}^{\, \cos x}\right)}^{\ln x} \left[ -\frac{1}{2}\sin x (\ln x)^2 + \frac{\cos x \ln x}{x} \right] \quad (x>0) $ |

$\displaystyle \left[ \sqrt{x}^{\, ({\cos x}^{\, \ln x} )} \right]’ = $ you know |

All right! Here marks the end of another relatively-short, 3000-words-ish *derivative module* to you. Hey, interested in launching a petition calling for the inclusion of **Exponent Rule** into your textbook? Or maybe you want a petition precisely to prevent this *absolute horror* from ever materializing? Either way, why not let us know by dropping a comment below, or by voicing your appraisal/opposition on Facebook or Twitter! 🙂

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